1
$\begingroup$

Let $z=(x,y) \in \mathbb{R}^N \times (0,+\infty)$, and let $$ P_m(z)=y^{2s} |z|^{-\frac{N+2s}{2}} K_{\frac{N+2s}{2}}(m|z|), $$ where $N \geq 3$ is an integer, $0<s<1$ and $K_{\frac{N+2s}{2}}$ denotes the modified Bessel function of the second kind with order $\frac{N+2s}{2}$.

For $u \in H^s(\mathbb{R}^N)$ and $$ v(x,y)=\int_{\mathbb{R}^N} P_m(x-z,y)u(z)\, dz, $$ I would like to show that $v$ decays exponentially fast as $|z|=\sqrt{|x|^2+y^2} \to +\infty$. My idea is to exploit the exponential behavior of $K$ at infinity, i.e. $$ K_{\frac{N+2s}{2}} (r) \sim r^{-1/2}e^{-r} \quad\hbox{when $r \to +\infty$}, $$ but I am unsure if this is actually correct. This question is related to the exponential behavior of solution to the so-called Caffarelli-Silvestre extension method for the operator $(-\Delta+m^2)^s$: $$ \left\{ \begin{array}{ll} -\operatorname{div}(y^{1-2s} \nabla v)+m^2 y^{1-2s}v=0 &\hbox{in $\mathbb{R}^N \times (0,+\infty)$} \\ -\lim_{y \to 0+} y^{1-2s} \frac{\partial v}{\partial y} = (-\Delta+m^2)^s u &\hbox{in $\mathbb{R}^N$}. \end{array} \right. $$ My conjecture is that $$ |v(x,y)| \leq c_1 e^{-c_2 y} e^{-c_3 \sqrt{|x|^2+y^2}} $$ for some constants $c_i>0$, $i=1,2,3$.

$\endgroup$
5
  • $\begingroup$ If $u(x)$ decays subexponentially, then $v(x,y)$ is roughly comparable to $u(x)$ as $|x| \to \infty$ (with $y$ fixed). Do we assume anything about the behaviour of $u(x)$ for large $x$? $\endgroup$ May 7, 2018 at 18:34
  • $\begingroup$ At least we can assume that $\lim_{|x| \to +\infty} u(x)=0$. $\endgroup$
    – Siminore
    May 8, 2018 at 7:13
  • $\begingroup$ This is not sufficient: it is not very hard to prove that if $u(x) \approx (1+|x|)^{-p}$, then $v(x,y) \approx (1+|x|)^{-p}$ for every fixed $y > 0$. By the way, I believe there is a typo in your expression for $P_m$: $\tfrac{N+2s}{2}$ should read $\tfrac{N-2s}{2}$, right? $\endgroup$ May 8, 2018 at 8:24
  • $\begingroup$ No, it seems to me that the numbers are correct. In a weaker way, I would be happy to show that eigenfunctions of the operator must decay exponentially fast in both variables. $\endgroup$
    – Siminore
    May 8, 2018 at 11:16
  • $\begingroup$ Regarding the exponents: you are right, I got confused by the Bessel-potential tag. $\endgroup$ May 8, 2018 at 19:35

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.